Q13E - University Physics with Modern Physics

Question

Two very narrow slits are spaced $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{ }\mathit{\mu }\mathit{m}$ apart and are placed $\mathbf{35}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$ from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with $\mathit{\lambda }\mathbf{=}\mathbf{550}\mathit{n}\mathit{m}$? (Hint: The angle $\mathit{\theta }$ in Eq. ( 35.5 ) is not small.)

Step-by-Step Solution

Answer

The distance between the first and second dark lines of the interference pattern is $\mathbf{12}\mathbf{.}\mathbf{6}\mathit{c}\mathit{m}$.

Step 1: Formula for distance between central bright fringe to the dark fringe

                                                                               $\mathit{y}\mathbf{=}\mathit{R}\mathbf{ }\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }$

Step 2: Calculate the angle for the first and second dark fringe

Given:            $\mathit{d}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{ }\mathit{\mu }\mathit{m}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathit{m}$

              $$\begin{gathered} \mathit{R}\mathbf{=}\mathbf{35}\mathit{c}\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathit{m} \\ \mathit{\lambda }\mathbf{=}\mathbf{550}\mathit{n}\mathit{m}\mathbf{=}\mathbf{550}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{ }\mathit{m} \end{gathered}$$           

The first dark fringe is for $\mathit{m}\mathbf{=}\mathbf{0}$ and the second is for $\mathit{m}\mathbf{=}\mathbf{1}$.

For dark fringes,

                                               $\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{(}\mathit{m}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{)}\mathit{\lambda }$      

Hence, 

                                                $\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\frac{\mathbf{(}\mathbf{m}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}\mathbf{\lambda }}{\mathbf{d}}$

                                                                                                                                                   (1)

The first dark fringe, when $\mathit{m}\mathbf{=}\mathbf{0}$

                                                    $\mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{\mathbf{(}\mathbf{0}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{)}\mathbf{\lambda }}{\mathbf{d}}\mathbf{\right]}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{\lambda }}{\mathbf{d}}\mathbf{\right]}$   

Plug the given,

                                                    $\mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{\mathbf{\left(}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\left(550*{10}^{-9}\right)\mathbf{\right)}\mathbf{\lambda }}{\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}}\mathbf{\right]}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{79}\mathbf{°}$

Similarly, for second dark fringe,

                                                                $\mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{\left(1.5\right)\mathbf{\lambda }}{\mathbf{d}}\mathbf{\right]}$

Plug the given,

                                                   $\mathbf{\Rightarrow }{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{*}\mathbf{550}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{)}}{\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}}\mathbf{\right]}\mathbf{=}\mathbf{27}\mathbf{.}\mathbf{3}\mathbf{°}$

Step 3: Calculate the distance between the first and second dark lines of the interference pattern

Now, the distance from the central bright fringe is given as:

                                                                          $\mathit{y}\mathbf{=}\mathit{R}\mathbf{ }\mathbf{tan}\mathit{\theta }$

Hence,

                                                            ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{R}\mathbf{ }\mathbf{tan}{\mathit{\theta }}_{\mathbf{1}}$ and ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathit{R}\mathbf{ }\mathbf{tan}{\mathit{\theta }}_{\mathbf{2}}$

So,

                                                        $\mathbf{∆}\mathit{y}\mathbf{=}{\mathit{y}}_{\mathbf{2}}\mathbf{-}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{R}\mathbf{ }\mathbf{tan}{\mathit{\theta }}_{\mathbf{2}}\mathbf{-}\mathit{R}\mathbf{ }\mathbf{tan}{\mathit{\theta }}_{\mathbf{1}}$

Plug the values,

                                               $\mathbf{∆}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{[}\mathbf{tan}\mathbf{27}\mathbf{.}\mathbf{3}\mathbf{°}\mathbf{-}\mathbf{tan}\mathbf{8}\mathbf{.}\mathbf{79}\mathbf{°}\mathbf{]}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{6}\mathit{c}\mathit{m}$

Thus, the distance between the first and second dark lines of the interference pattern is $\mathbf{12}\mathbf{.}\mathbf{6}\mathit{c}\mathit{m}$.